Solution Manual-Engineering Circuit Analysis
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
1.
(a) 12 μs (b) 750 mJ (c) 1.13 kΩ
(d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz
(g) 39 pA (h) 49 kΩ (i) 11.73 pA
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
2.
(a) 1 MW (b) 12.35 mm (c) 47. kW (d) 5.46 mA
(e) 33 μJ (f) 5.33 nW (g) 1 ns (h) 5.555 MW
(i) 32 mm
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
3.
(a)
( 400 Hp ) ⎜
⎛ 745.7 W ⎞ ⎟ = ⎝ 1 hp ⎠
298.3 kW
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ (b) 12 ft = (12 ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.658 m ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ (c) 2.54 cm = ⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = ⎝ 1 Btu ⎠ (e) 285.4´10-15 s = 25.4 mm 70.69 kJ 285.4 fs
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
4.
(15 V)(0.1 A) = 1.5 W = 1.5 J/s. 3 hrs running at this power level equates to a transfer of energy equal to (1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
Chapter Two Solutions
10 March 2006
1.
(a) 12 μs (b) 750 mJ (c) 1.13 kΩ
(d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz
(g) 39 pA (h) 49 kΩ (i) 11.73 pA
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
2.
(a) 1 MW (b) 12.35 mm (c) 47. kW (d) 5.46 mA
(e) 33 μJ (f) 5.33 nW (g) 1 ns (h) 5.555 MW
(i) 32 mm
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
3.
(a)
( 400 Hp ) ⎜
⎛ 745.7 W ⎞ ⎟ = ⎝ 1 hp ⎠
298.3 kW
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ (b) 12 ft = (12 ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.658 m ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ (c) 2.54 cm = ⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = ⎝ 1 Btu ⎠ (e) 285.4´10-15 s = 25.4 mm 70.69 kJ 285.4 fs
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions
10 March 2006
4.
(15 V)(0.1 A) = 1.5 W = 1.5 J/s. 3 hrs running at this power level equates to a transfer of energy equal to (1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition
Chapter Two Solutions